Question

Beer shelf life is a problem for brewers and distributors because when beer is stored at room temperature, its flavor deteriorates. When the average furfuryl ether content reaches 6 μg per liter, a typical consumer begins to taste an unpleasant chemical flavor. At α = 0.05, would the following sample of 12 randomly chosen bottles stored for a month convince you that the mean furfuryl ether content exceeds the taste threshhold?
6.94 7.50 5.87 7.92 8.66 7.78 7.95 6.35 6.16 6.28 5.68 7.02
For the following hypothesis:
a) H0: μ ≤ 6 versus H1: μ > 6, what is the value of the test statistic?
b) What is the p-value?
c) At α = .05, we would
i. reject the null hypothesis.
ii.fail to reject the null hypothesis.

Answer 1

Answer:

a) 3.52

b)  0.002399

c)  At α = .05 we reject the null hypothesis.

Step-by-step explanation:

We are given the following information:

Population mean, μ = 6 μg per liter

Sample size, n = 12

Alpha, α = 0.05

6.94, 7.50, 5.87, 7.92, 8.66, 7.78, 7.95, 6.35, 6.16, 6.28, 5.68, 7.02

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{84.11}{12} = 7.009$

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$SS.D = \sqrt{\frac{10.03}{11}} = 0.95$

First, we design the null and the alternate hypothesis

$H_{0}: \mu \leq 6\\H_A: \mu > 6$

We use One-tailed t test to perform this hypothesis.

a) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{7.009 - 6}{\frac{0.95}{\sqrt{11}} } = 3.52$

b) P-value for one tailed t-test at 11 degree of freedom and α = 0.05 is 0.002399

c) Since P-value < 0.05

The result is insignificant.

Thus, at α = .05 we reject the null hypothesis.